Abstract:
Burning ships, perfect lenses, Red shirts
and weird Al
Hello again folks...
Just for spite. this post is just for spite.
well, not really but it's fun to say it is.
As some of you might remember I wrote an inhumanely long post, and my liege-wife told me to cut it short.
so I did, but no good deed goes unpunished or as theorized by the great-but-not-overly-sane Newton "To every action there is always opposed an equal reaction", and so it is with great yet perverse pleasure, that I give you the second part of the Lenses & Idiots Trilogy(!!!)
 |
| Not quite the shilling, though he probably minted some shillings... |
Anyway, last time I was reminded of two nice stories....
The first about Archimedes and the Siege of Syracuse, where supposedly the defending forces used about 300 round bronze shields, polished to a shine, in order to concentrate sunlight, and burn the marauding fleet to a crisp.did anyone say he loved the smell of napalm in the morning?
well let's put that theory to the test:
groceries:
1. 300 bronze body shields of area 1.3m x 0.7 m, polished to an "\(\varepsilon\)" sheen.
2. the sun - Earth's solar constant. - 1360 \(\frac{J}{sec\cdot m^2}\)
3. auto-combustion temperature of wood - at the most \(~ 450^{\circ}c\)
4. a distance estimate say at 500 meter.
and now:
300 shields yield an incident area of about 273 square meters. thus according to our previous calculations we simply substitute the respective areas and get the temperature at a 1 square meter of surface on the wood to be about 1600K at \(\varepsilon=1\), which means this should burn like kindle wood but experimental results here assert differently... so what went wrong?
well, a couple of things:
first off, bronze is not a mirror, meaning for starters, \(\varepsilon\neq1\) in fact not even close.
secondly, we assumed the image area of the shields on the side of the boat is at the same size of the shield. this is miserably wrong.
Consider a piece of shiny metal or a wrist-watch you use to blind the lecturer at a "Mechanics and Special Relativity" course, it is pretty obvious that the area of light traced by the image of the watch's surface is larger than the actual surface of the watch right?
The added length of a line image is given by
\[L_{image-line}=L_{original-line}\left(1+\sin(\alpha)r\right)\]
so basically the area increases like
\[A=A_{original}\left(1+2\sin(\alpha)r+\sin^2(\alpha)r^2\right)\]
So even though the flux is multiplied 300-fold, by all the shields reflecting the sun at the same site, still the area of incident is also multiplied, and even if we take only the linear approximation and not the whole deal we STILL get the incident area to grow like \(1+cr\) thus:
\[\Phi_{boat}=1360\cdot 300 \cdot \frac{1}{1+c500}\]
And even taking the angle to be fairly small let's say \(15^{\circ}\) we get the temperature to be about 410K which is about \(110^{\circ}c\) and that with an \(\varepsilon\) of 1!! its enough that we assume \(\varepsilon\) to be 50% which is of course a gross over estimate, we get a temperature of about \(50^{\circ}c\).
That kind of a temperature is not enough to boil water, much less trigger wood auto-combustion ,in fact, even if we were talking about perfect reflective surfaces at vacuum conditions this just ain't enough or differently put: "I just can't do it captain, I don't have the power" - Hmm... maybe if the wood was treated with a combustion agent first...
So the obvious conclusion here is that the people of Syracuse had an inside agent!!!
They had a traitor in their midst!!!!
Or most likely this never actually happened....
Warning- computer geek humor below:
<Computer geek humor>
|
|
By the way, did you ever notice that while redshirts are being slaughtered by the dozens in the original star trek series, Scotty actually wears a red shirt, but is immune to the redshirt-death-rule?
hmm... makes me want to throw an exception....
</Computer geek humor>
Anyway...
Let's see what happens when we apply a lens at the 1 square meter target area, with such accuracy as to concentrate the rays at an area no bigger than 1 square centimeter, we get with the same type of calculations, even taking into account the \(\frac{1}{r}\) factor, a temperature of about 4100K, and if we have for example a solar tower, surrounded by perfectly reflective parabolic mirrors, that span an area of,say, 500 square meters, all directed at a Zeiss Parabolic lens of a perfect nature, with the longest mirror to lens distance of about 200 meters, and an \(\varepsilon\) factor of 70% we get, about 5400K.
Again what melts at 5400K?
hmm a short list that includes about everything...
a short list of things that actually BOIL at 5400K includes among others:
Carbon, Platinum, Rhodium, Titanium, Silicon, Palladium, Cobalt, Nickel, Iron etc...
For a reference the surface temperature of the SUN is evaluated at about 5800K.
But, interestingly enough, a star-ship that tries to show-off and make a run near the sun will have to withstand the ludicrously high temperature of about 5 million Kelvin of the sun's corona, about 3 orders of magnitude higher then the surface temperature - thus while bathing in the sun's surface sea of fire might be enjoyable, getting there could prove messy.
Of course this goes against star-trek TNG's episode "redemption II" story-line, when a Klingon Bird Of Prey survived the corona only to explode on the surface...
 |
| A Klingon BOP taking a nice warm bath in a sun. |
The second story goes something like this:
I have a friend who's father is the head of solar energy research at a notable institute.
He told me the story of acquiring an almost perfect Zeiss lens. and it goes something like this:
One day, he and one of his colleagues were wondering the streets of Dresden Germany, biding their time in between lectures, when they saw a group of kinder-garden age kids, playing around with a large lens, having fun with reflections and images.
This physics professor immediately recognized the lens to be of tremendous quality, and approached the kinder-garden staff with an offer to buy the lens.
They really didn't know what they had in their hands, or simply didn't care too much, but they pretty much GAVE the lens away to that professor, with a simple demand - give us a lens that will do what this one know how to do, so the kids will be able to continue playing. He gave them an ordinary, albeit good quality lens, and basically got this magnificent lens for a song...
It is that lens that is still on the top of the solar tower at his laboratory, and through which they get temperatures high enough to burn through steel as if it was butter.
Now as promised a short mass on Fresnel Lenses vs. parabolic lenses:
In short - a parabolic lens has the interesting feature where every light beam coming straight from infinity hits the same focal point, and vise-verse, if you put a light source at the focal point of a perfect parabolic lens, you can be damn sure that all the light goes straight ahead - in fact that is how the high beams on your car works, in other words - know the mechanism of the bastard that blinds you!!!
To show that, we take a simple parabola - \(f(x)=\alpha x^2\), and we will take a beam that comes from straight up and hits the parabola at \(x=x_0\).
the tangent to the graph at that point is given by \(y=2\alpha x_0 x+b\), and the beam that hits the parabola at that point is diverted to a line that looks like \(y=-\frac{1-(2\alpha x_0)^2}{4\alpha x_0}x +c\) Where calculating \(c\) yields \(c=\frac{1}{4\alpha}\).
The fact that where the diverted light ray cuts the "y" axis, doesn't include \(x_0\) as an argument already shows that all rays meet at the same place i.e. at the same focal point at \(\left(0,\frac{1}{4\alpha}\right)\).
A Fresnel lens is a neat way to manufacture a closely approximated parabolic lens, while reducing physical bulk and bill of materials. What you want to do is take a regular parabolic lens, slice it in regions, and piece together what you got. I found a nice illustration of this on the net, here is the picture...
 |
| Fresnel Vs. Parabolic lens, notice the corresponding regions. |
Thus it is pretty clear, that while Parabolic lenses capture all light rays, it is heavier and bulkier than the corresponding Fresnel lens.
On the other hand Fresnel lenses capture MOST light ray, but not all, so while being lighter and more compact, it creates some distortion in the image, depending of course on the quality of material used, and the "resolution" or the density of "cut&paste" done to the original Parabolic lens to get the Fresnel lens.
Of course this is a very low concern for lighthouses, and other navigational signs and lights, like for instance masthead, port and starboard lights and also port entry and hazard lights.
And so we see that using Fresnel lenses is more efficient in the long run, because it takes up more energy to rotate or move a bulky and cumbersome lens then a lightweight one, and of course reducing weight and energy immediately corresponds to corrosion and mechanical faults ratio reduction.
So why not use a parabolic reflective surface? this could be lightweight as well as cheap AND efficient in terms of light reflection?
Well I think today most mechanisms where we have enough space to imbed that surface, actually DO use reflective parabolic surfaces, another good example of this would probably be the lenses and mirrors inside a telescope.
As for why NOT to use parabolic lenses and mirrors? - mainly physical space considerations probably, but also, in older parabolic mirrors I THINK there might have been significant energy losses via a nifty little mechanism called the "skin effect" but that is all for now, I may discuss this effect in a different post.
Wow, this was long, I hope you enjoyed this...
If indeed you have, you have the making of a true nerd. Oh and sorry I didn't get to almost die in this post, I have plenty of cases in which I did, and I hope they will always continue to be of the "almost" type in order for me to continue recounting them...
 |
| a true nerd... |
As always, next time : oh... why bother...
